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Diagonal sums of the number triangle associated to A086617.
+20
2
1, 1, 2, 3, 5, 8, 14, 24, 43, 78, 144, 269, 509, 971, 1868, 3618, 7049, 13805, 27162, 53661, 106405, 211697, 422458, 845386, 1696017, 3410522, 6873060, 13878721, 28077439, 56900936, 115501012, 234807488, 478032437, 974507543, 1989123814
FORMULA
G.f.: (1-x^2-sqrt(1-2*x^2-4*x^3-3*x^4))/(2*x^3*(1-x^2)).
a(n) = Sum_{k=0..floor(n/2)} Sum_{j=0..n-2*k} binomial(n-2*k, j)*binomial(k, j) * A000108(j). Conjecture: (n+3)*a(n) +(-n-2)*a(n-1) +2*(-n-1)*a(n-2) +2*(-n+3)*a(n-3) +(n+1)*a(n-4) +3*(n-2)*a(n-5)=0. - R. J. Mathar, Nov 16 2012
Antidiagonal sums of triangle A086614.
+10
19
1, 2, 4, 8, 17, 38, 89, 216, 539, 1374, 3562, 9360, 24871, 66706, 180340, 490912, 1344379, 3701158, 10237540, 28436824, 79288843, 221836402, 622599625, 1752360040, 4945087837, 13988490338, 39658308814, 112666081616
COMMENTS
a(n) is the number of distinct ordered trees obtained by branch-reducing the ordered trees on n+1 edges. - David Callan, Oct 24 2004 a(n) is the number of consecutive horizontal steps at height 0 of all Motzkin paths from (0,0) to (n,0) starting with a horizontal step. - Charles Moore (chamoore(AT)howard.edu), Apr 15 2007
This sequence (with offset 1 instead of 0) occurs in Section 7 of K. Grygiel, P. Lescanne (2015), see g.f. N. - N. J. A. Sloane, Nov 09 2015 Also number of plain (untyped) normal forms of lambda-terms (terms that cannot be further beta-reduced.) [Bendkowski et al., 2016]. - N. J. A. Sloane, Nov 22 2017 If interpreted with offset 2, the INVERT transform is A002026 with offset 1. - R. J. Mathar, Nov 02 2021
FORMULA
G.f.: A(x) = 1/(1-x)^2 + x^2*A(x)^2.
a(n) = Sum_{k=0..floor((n+1)/2)} binomial(n+1, 2k+1)*binomial(2k, k)/(k+1). - Paul Barry, Nov 29 2004 a(n) = n + 1 + Sum_k a(k-1)*a(n-k-1), starting from a(n)=0 for n negative. - Henry Bottomley, Feb 22 2005 a(n) = Sum_{k=0..n} Sum_{j=0..n-k} C(j)*C(n-k, 2j). - Paul Barry, Aug 19 2005 G.f.: c(x^2/(1-x)^2)/(1-x)^2, c(x) the g.f. of A000108; a(n) = Sum_{k=0..floor(n/2)} C(n+1,n-2k)*C(k). (End)
Binomial transform of doubled Catalan sequence 1,1,1,1,2,2,5,5,14,14,... - Paul Barry, Nov 17 2005 g(z) = (1-z-sqrt(1-2z-3z^2))/(2z-2z^2)/z - Charles Moore (chamoore(AT)howard.edu), Apr 15 2007, corrected by Vaclav Kotesovec, Feb 13 2014 D-finite with recurrence (n+2)*a(n) +3*(-n-1)*a(n-1) +(-n+4)*a(n-2) +3*(n-1)*a(n-3)=0. - R. J. Mathar, Nov 30 2012
EXAMPLE
a(0)=1, a(1)=2, a(2)=3+1=4, a(3)=4+4=8, a(4)=5+10+2=17, a(5)=6+20+12=38, are upward antidiagonal sums of triangle A086614: {1},
{2,1},
{3,4,2},
{4,10,12,5},
{5,20,42,40,14},
{6,35,112,180,140,42}, ...
For example, with n=2, the 5 ordered trees ( A000108) on 3 edges are |...|..../\.../\.../|\..
|../.\..|......|........
|.......................
Suppressing nonroot vertices of outdegree 1 (branch-reducing) yields
|...|..../\.../\../|\..
.../.\.................
of which 4 are distinct. So a(2)=4.
a(4)=8 because we have HHHH, HHUD, HUDH, HUHD
MAPLE
option remember;
if n <= 3 then
2^n;
else
3*(-n-1)*procname(n-1) +(-n+4)*procname(n-2) +3*(n-1)*procname(n-3) ;
-%/(n+2) ;
end if;
end proc:
MATHEMATICA
CoefficientList[Series[(1-x-Sqrt[1-2*x-3*x^2])/(2*x-2*x^2)/x, {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 13 2014 *)
a(n) = Sum{k=0..n} binomial(n,k)^2*C(k), where C() = A000108() are the Catalan numbers.
+10
11
1, 2, 7, 33, 183, 1118, 7281, 49626, 349999, 2535078, 18758265, 141254655, 1079364105, 8350678170, 65298467487, 515349097713, 4100346740511, 32858696386766, 265001681344569
COMMENTS
Main diagonal of square table A086617 of coefficients, T(n,k), of x^n*y^k in f(x,y) that satisfies f(x,y) = 1/[(1-x)(1-y)] + xy*f(x,y)^2. a(n) is the number of permutations of length 2n which are invariant under the reverse-complement map and have no decreasing subsequences of length 4. - Eric S. Egge, Oct 21 2008 In 2012, Zhi-Wei Sun proved that for any odd prime p we have the congruence a(1) + ... + a(p-1) == 0 (mod p^2). - Zhi-Wei Sun, Aug 22 2013
FORMULA
Recurrence: (n+3)^2*(4*n+7)*a(n+2) = 2*(20*n^3+117*n^2+220*n+135)*a(n+1) - 9*(n+1)^2*(4*n+11)*a(n). - Vaclav Kotesovec, Sep 11 2012 G.f.: (1-(1-9*x)^(1/3)*hypergeom([1/3,1/3],[1],-27*x*(1-x)^2/(1-9*x)^2))/(6*x). - Mark van Hoeij, May 02 2013 D-finite with recurrence (n+1)^2*a(n) +(-19*n^2+8*n+6)*a(n-1) +9*(11*n^2-30*n+21)*a(n-2) -81*(n-2)^2*a(n-3)=0. - R. J. Mathar, Aug 01 2022
EXAMPLE
a(5) = 1118 = 1*1^2 + 1*5^2 + 2*10^2 + 5*10^2 + 14*5^2 + 42*1^2.
MATHEMATICA
Flatten[{1, RecurrenceTable[{(n+3)^2*(4*n+7)*a[n+2]==2*(20*n^3+117*n^2+220*n+135)*a[n+1]-9*(n+1)^2*(4*n+11)*a[n], a[1]==2, a[2]==7}, a, {n, 1, 20}]}] (* Vaclav Kotesovec, Sep 11 2012 *) Table[HypergeometricPFQ[{1/2, -n, -n}, {1, 2}, 4], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 03 2016 *)
PROG
(PARI) a(n)=sum(k=0, n-1, (4*k+3)*sum(i=0, k, binomial(k, i)^2*binomial(2*i, i)))/3/n^2 \\ Charles R Greathouse IV, Sep 12 2012
Number of odd length roads between any adjacent nodes in virtual optimal chordal ring of degree 3 (the length of chord < number of nodes/2).
+10
4
1, 5, 31, 213, 1551, 11723, 90945, 719253, 5773279, 46889355, 384487665, 3177879675, 26442188865, 221278343445, 1860908156031, 15717475208853, 133256583398655, 1133591857814363, 9672323357640129, 82752014457666363, 709719620585186529, 6100394753270329605
REFERENCES
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, 2nd ed. 1998, see page number?
FORMULA
a(1) = 1; a(n) = 9*a(n-1) - 2* A086618(n), where A086618(n) = Sum_{k=0..n} Catalan(n)*binomial(n, k)^2, and Catalan(n) = (2*n)!/(n!*(n+1)!). - Michael SomosRecurrence: n^2*a(n) = (10*n^2-10*n+3)*a(n-1) - 9*(n-1)^2*a(n-2). - Vaclav Kotesovec, Oct 14 2012 G.f.: (hypergeom([1/3, 1/3],[1],-27*x*(x-1)^2/(9*x-1)^2)/(1-9*x)^(2/3)-1)/3. - Mark van Hoeij, May 14 2013 G.f.: G(0)/(6*x*(1-9*x)^(2/3) ) -1/(3*x), where G(k)= 1 + 1/(1 - 3*(3*k+1)^2*x*(1-x)^2/(3*(3*k+1)^2*x*(1-x)^2 - (k+1)^2*(1-9*x)^2/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 31 2013 a(n) = hypergeom([1/2, -n, -n], [1, 1], 4) / 3. - Peter Luschny, Nov 06 2023
EXAMPLE
a(1)=1; a(2)=9*a(1)-2*2=9-4=5; a(3)=9*5-2*7=31; a(4)=9*31-2*33=213; etc
MAPLE
a := 1; s := 0; for k from 1 to 10 do for i from 0 to k do ss := ((2*(i))!/((i)!*(i+1)!))*((k)!/((i)!*(k-i)!))^2; s := s+ss; od; a := (9*a-2*s); s := 0; od;
# Alternative:
a := n -> hypergeom([1/2, -n, -n], [1, 1], 4)/3;
MATHEMATICA
Table[Sum[Binomial[n, k]^2*Binomial[2k, k], {k, 0, n}]/3, {n, 1, 20}] (* Vaclav Kotesovec, Oct 14 2012 *)
PROG
(PARI) a(n) = sum(k=0, n, binomial(n, k)^2*binomial(2*k, k))/3; \\ Michel Marcus, May 10 2020
AUTHOR
B. Dubalski (dubalski(AT)atr.bydgoszcz.pl), Oct 23 2003
Square table, read by antidiagonals, of coefficients T(n,k) of x^n*y^k in f(x,y) that satisfies f(x,y) = 1/(1-x-y) + xy*f(x,y)^3.
+10
3
1, 1, 1, 1, 3, 1, 1, 6, 6, 1, 1, 10, 21, 10, 1, 1, 15, 55, 55, 15, 1, 1, 21, 120, 212, 120, 21, 1, 1, 28, 231, 644, 644, 231, 28, 1, 1, 36, 406, 1652, 2617, 1652, 406, 36, 1, 1, 45, 666, 3738, 8685, 8685, 3738, 666, 45, 1, 1, 55, 1035, 7680, 24735, 36345, 24735, 7680
COMMENTS
The g.f. for A001764 satisfies: g(x) = 1 + x*g(x)^3.
EXAMPLE
Rows begin:
{1, 1, 1, 1, 1, 1, 1, 1,..}
{1, 3, 6,10,15,21,28,..}
{1, 6,21,55,120,231,..}
{1,10,55,212,644,..}
{1,15,120,644,..}
{1,21,231,..}
MATHEMATICA
t[n_, k_] := Sum[ Binomial[n+k, 2*i]*Binomial[n+k-2*i, k-i]*(3*i)!/(i!*(2*i+1)!), {i, 0, k}]; Table[t[n-k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 18 2013, after Michael Somos *)
Symmetric square table of coefficients, read by antidiagonals, where T(n,k) is the coefficient of x^n*y^k in f(x,y) that satisfies: f(x,y) = g(x,y) + xy*f(x,y)^4 and where g(x,y) satisfies: 1 + (x+y-1)*g(x,y) + xy*g(x,y)^2 = 0.
+10
3
1, 1, 1, 1, 4, 1, 1, 10, 10, 1, 1, 20, 48, 20, 1, 1, 35, 162, 162, 35, 1, 1, 56, 441, 841, 441, 56, 1, 1, 84, 1036, 3314, 3314, 1036, 84, 1, 1, 120, 2184, 10786, 18004, 10786, 2184, 120, 1, 1, 165, 4236, 30460, 77952, 77952, 30460, 4236, 165, 1, 1, 220, 7689, 77044
COMMENTS
Explicitly, g(x,y) = ((1-x-y)+sqrt((1-x-y)^2-4xy))/(2xy) = sum(n>=0, sum(k>=0, N(n,k)*x^n*y^k), where N(n,k) are the Narayana numbers: N(n,k) = C(n+k,k)*C(n+k+2,k+1)/(n+k+2). This array is directly related to sequence A002293, which has a g.f. h(x) that satisfies h(x) = 1 + x*h(x)^4. The inverse binomial transform of the rows grows by three terms per row.
EXAMPLE
Rows begin:
[1, 1, 1, 1, 1, 1, 1, 1, ...];
[1, 4, 10, 20, 35, 56, 84, 120, ...];
[1, 10, 48, 162, 441, 1036, 2184, 4236, ...];
[1, 20, 162, 841, 3314, 10786, 30460, 77044, ...];
[1, 35, 441, 3314, 18004, 77952, 284880, 912042, ...];
[1, 56, 1036, 10786, 77952, 435654, 2007456, 7951674, ...];
[1, 84, 2184, 30460, 284880, 2007456, 11427992, 55009548, ...];
[1, 120, 4236, 77044, 912042, 7951674, 55009548, 317112363, ...];
[1, 165, 7689, 178387, 2624453, 27870393, 231114465, 1576219474, ...]; ...
PROG
(PARI) {L=10; T=matrix(L, L, n, k, 1); for(n=1, L-1, for(k=1, L-1, T[n+1, k+1]=binomial(n+k, k)*binomial(n+k+2, k+1)/(n+k+2)+ sum(j3=1, k, sum(i3=1, n, T[n-i3+1, k-j3+1]* sum(j2=1, j3, sum(i2=1, i3, T[i3-i2+1, j3-j2+1]* sum(j1=1, j2, sum(i1=1, i2, T[i2-i1+1, j2-j1+1]*T[i1, j1])); )); )); )); T}
Triangle read by rows, where T(n,k) is the coefficient of x^n*y^k in f(x,y) that satisfies f(x,y) = 1/(1-x)^2 + xy*f(x,y)^2.
+10
2
1, 2, 1, 3, 4, 2, 4, 10, 12, 5, 5, 20, 42, 40, 14, 6, 35, 112, 180, 140, 42, 7, 56, 252, 600, 770, 504, 132, 8, 84, 504, 1650, 3080, 3276, 1848, 429, 9, 120, 924, 3960, 10010, 15288, 13860, 6864, 1430, 10, 165, 1584, 8580, 28028, 57330, 73920, 58344, 25740
FORMULA
T(n,k) = binomial(2*k, k-1)*binomial(n+k+1, n-k) / k for k > 0. # Peter Luschny, Jan 26 2018
EXAMPLE
Rows:
{1},
{2, 1},
{3, 4, 2},
{4, 10, 12, 5},
{5, 20, 42, 40, 14},
{6, 35, 112, 180, 140, 42},
{7, 56, 252, 600, 770, 504, 132},
{8, 84, 504, 1650, 3080, 3276, 1848, 429}, ...
MAPLE
T := (n, k) -> `if`(k=0, n+1, binomial(2*k, k-1)*binomial(n+k+1, n-k)/k):
for n from 0 to 8 do seq(T(n, k), k=0..n) od; # Peter Luschny, Jan 26 2018
Triangular sequence based on Pascal's triangle: t(n,m) = 2*binomial(m, n) - (1 + n*(m - n)).
+10
0
1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 7, 4, 1, 1, 5, 13, 13, 5, 1, 1, 6, 21, 30, 21, 6, 1, 1, 7, 31, 57, 57, 31, 7, 1, 1, 8, 43, 96, 123, 96, 43, 8, 1, 1, 9, 57, 149, 231, 231, 149, 57, 9, 1, 1, 10, 73, 218, 395, 478, 395, 218, 73, 10, 1
COMMENTS
Suggested by Gary W. Adamson from a previous submission. Very close to (but slightly smaller at 7th row) A086617.
FORMULA
t(n,m) = 2*binomial[m, n] - (1 + n*(m - n)).
EXAMPLE
{1},
{1, 1},
{1, 2, 1},
{1, 3, 3, 1},
{1, 4, 7, 4, 1},
{1, 5, 13, 13, 5, 1},
{1, 6, 21, 30, 21, 6, 1},
{1, 7, 31, 57, 57, 31, 7, 1}
MATHEMATICA
Table[Table[2*Binomial[m, n] - (1 + n*(m - n)), {n, 0, m}], {m, 0, 10}] Flatten[%]
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